# Tag Info

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Right so what you have misunderstood is that this was a statistical prediction about when the first detection might be made, based on assumptions about the number of possible sources in the universe and their distance from us. It is mildly surprising that (advanced) LIGO found such a big signal so soon after it began operations. Detecting a few more will ...

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About half of the proton's momentum is carried by the valence quarks (uud) and the rest is spread around many gluons and many sea-quarks ($q\bar{q}$ pairs). In the Drell-Yan process, it is assumed that the colliding quark is a valence quark and the anti-quark must be a sea quark. We are colliding a highly energetic quark with a low energy anti-quark. This ...

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A non-quantam mechanic answer - would be electrostatic repulsions (b/w -vely charged electrons) would prevent it and give them definite paths which don't intersect. Even if they move with 0.3c they can't comes much nearer than 4 fm.

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If the collision is inelastic, momentum is conserved, but energy is not. Some of the energy is converted to heat (and sound).

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There is a general definition of a center of mass in General Relativity, which was proposed by Dixon (W.G. Dixon, Il Nuovo Cimento 34, 317 (1964)) and whose existence and uniqueness were proved by Beiglböck (W. Beiglbock, Commun. Math. Phys. 5, 106 (1967)). These are very old papers, so that people have been working since long ago on this topic. In ...

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Yes. There are several mathematical models you can use to describe an elastic collision. One of the most popular is the hard-sphere potential which assumes that the molecules are spheres of radius $a$ that cannot interpenetrate each other. Mathematically, it takes the form $V(r) = 0$ if $r>2a$ $V(r) = +\infty$ if $r<2a$, $r$ being the distance ...

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So this is an oversimplification of the impact process, but if you model it as a spring damper system, then the deflection response is $$x(t) = X \exp(-\zeta \omega_n t) \sin (\omega_n t \sqrt{1-\zeta^2})$$ where \begin{align} \omega_n & =\sqrt{ \frac{k}{m}} && \mbox{: natural frequency, rad/s} \\ \zeta &= \frac{c}{2 m \omega_n} ...

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In Galilean relativity (pretty sure that is what you mean/need), you just add up the velocities. So $V_{p_1} =v/2+x$ and $V_{p_2}=v/2+y=v/2-x$

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Exactly. In a 2D problem, it's usually a good idea to break the components into two dimensions based on the environment. In this case, the wall makes the best split, let's say that x is the direction of motion along the wall while y is perpendicular. In this simple situation, the force on the ball can only act in the Y direction. Which has the following ...

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This assumes a smooth surface collision. The component of velocity (momentum) along the surface of the wall cannot change because there is no friction and hence no forces along that direction. Because the mass of the wall is assumed to be much greater that the mass of the ball and the collision is assumed to be elastic the normal component of velocity of the ...

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Parts of the masses must come to rest if one or both of the masses change direction. The system of two masses has a centre of mass with linear momentum $m_A u - m_B u$ in the direction of $A$'s travel before the collision. Since there are no external forces acting that momentum cannot change and so there must always be parts of those masses which are moving. ...

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This is essentially the same question as https://astronomy.stackexchange.com/questions/13302/ In a perfectly elastic collision, both momentum and kinetic energy are conserved. The initial momentum is $\text{m1} \text{v1}$ and the initial kinetic energy is $\frac{\text{m1} \text{v1}^2}{2}$, since m2 is at rest. Let u1 and u2 be the velocities of the ...

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A rod of mass $M$ and length $\ell$ has mass moment of inertia $I = \frac{M}{12} \ell^2$. The impact at a distance of $c = \frac{\ell}{2}$ from the center of mass imparts an impulse $J$, while an equal and opposite impulse $-J$ is applied to the projectile mass $m$. The projectile is going to bounce with velocity $v_B = v - \frac{J}{m}$. The center of mass ...

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The linear motion of the bars center in the direction of the ball will be as if the ball had hit the bar at its center. The ball will bounce off the bar in the exact opposite direction but with reduced speed. The ball will also make the bar rotate around its center with some angular velocity. After impact, the sum of the translational energy of the ball and ...

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Let us take the system - Rod + Ball Let the final velocity of center of mass of rod and ball be $v_1$ and $v_2$ respectively. As there is no net external force on the system, the net linear momentum of the system will be conserved. $mv = Mv_1 + mv_2$ And, as there is no net external torque about the COM of the system, the angular momentum of the ...

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This is a standard rotational motion problem. Use conservation of linear momentum for the translational motion. Use conservation of angular momentum about any axis noting that some axes make the algebra easier than others. Use conservation of kinetic energy as the collision is stated to be elastic.

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The acceleration at the point of reflection is actually quite complicated. It is caused by the elastic forces of the surface and the ball and has a complicated time dependence. However, the timespan in which the ball touches the ground is very short (especially if the ground and the ball are very rigid), therefore we can simplify the actual acceleration ...

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The behavior of the balls in Newton's cradle does provide a clue to understanding this phenomenon. Because of the spherical symmetry of the balls, and the fact that they make contact at a point, the propagation of compression pulses through the array is not like regular sound propagation through a solid medium. Sound in any medium is usually subject to ...

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The majority of the energy is dissipated as mechanical deformation (as Jon Custer has stated). Visualizing the situation can help a lot. All matter is somewhat elastic - there's no such thing as infinite elasticity. When an object collides with something, the force of the collision takes time to spread out across the object. Imagine a slow-motion view of ...

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Well let's say you weigh 220 lbs. Which translates over to 100 kg. The fall is 50 ft, so about 15.24 meters. The running thing that most people say is that it takes about 5000 Newtons of force to break a human bone, but we know that, this varies. We also know that it is not how hard you hit something that necessarily kills you, it's energy from the impact, ...

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The energy has several options to get dissipated into: Major part of it is turned into heat as a result of friction. Some part gets transmitted into sound energy, causing the sound we hear when the object falls. A very feeble amount gets transformed into light energy. Another miniscule portion is utilised in deforming the object and thereby increasing the ...

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It's not the fall that kills you. Your chance of survival is 100% :) In regards to the suspected impact after the fall, you will need to expand on the parameters to your question. Divers survive 50ft drops routinely. The elderly are killed from 4 foot drops routinely. I'm sure this wasn't the answer you were looking for, but I think the best answer ...

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Are talking about Newtonian (Classical) mechanics - Physics I and II? if so, in that case all you need are Conservation of Energy and Conservation of Momentum. You ignore external forces.

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